MHB Solve Equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$ in Real Numbers

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The equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$ is solved in real numbers, leading to the conclusion that $a=0$ is the only solution. A method involving the transformation $3a = 3\sqrt[3]{a(a^2-1)}$ was discussed, resulting in $a^3 = a(a^2-1)$. Initially, there was confusion regarding the solution, but it was clarified that $a=0$ is indeed correct. Participants acknowledged calculation errors in their methods. Ultimately, the consensus is that $a=0$ is the sole solution to the equation.
anemone
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Solve in real numbers the equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$
 
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anemone said:
Solve in real numbers the equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$

using

x+y+z=0=>$x^3+y^3+x^3 = 3xyz$

we get

$(3a)^3 = 3a(a^2-1)$
a= 0 or +/-$\sqrt(1/8)$
 
kaliprasad said:
using

x+y+z=0=>$x^3+y^3+x^3 = 3xyz$

we get

$(3a)^3 = 3a(a^2-1)$
a= 0 or +/-$\sqrt(1/8)$
Neat method, but I get a different answer.

[sp]From $3a = 3\sqrt[3]{a(a^2-1)}$, I get $a^3 = a(a^2-1)$, with $a=0$ the only solution.[/sp]
 
Opalg said:
Neat method, but I get a different answer.

[sp]From $3a = 3\sqrt[3]{a(a^2-1)}$, I get $a^3 = a(a^2-1)$, with $a=0$ the only solution.[/sp]

There was a calculation mistake in my method
 
Thanks to both of you for participating and yes, $a=0$ is the only answer to the problem.:)
 
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